已知函數(shù)f(x)=(a+1)lnx+ax2+1
(1)討論函數(shù)f(x)的單調(diào)性;
(2)設(shè)a<-1.如果對任意x1,x2∈(0,+∞),|f(x1)-f(x2)|≥4|x1-x2|,求a的取值范圍.
【答案】
分析:(1)先確定函數(shù)的定義域然后求導(dǎo)數(shù)fˊ(x),在函數(shù)的定義域內(nèi)解不等式fˊ(x)>0和fˊ(x)<0,求出單調(diào)區(qū)間.
(2)根據(jù)第一問的單調(diào)性先對|f(x
1)-f(x
2)|≥4|x
1-x
2|進行化簡整理,轉(zhuǎn)化成研究g(x)=f(x)+4x在(0,+∞)單調(diào)減函數(shù),再利用參數(shù)分離法求出a的范圍.
解答:解:(Ⅰ)f(x)的定義域為(0,+∞).

.
當a≥0時,f′(x)>0,故f(x)在(0,+∞)單調(diào)增加;
當a≤-1時,f′(x)<0,故f(x)在(0,+∞)單調(diào)減少;
當-1<a<0時,令f′(x)=0,解得

.
則當

時,f'(x)>0;

時,f'(x)<0.
故f(x)在

單調(diào)增加,在

單調(diào)減少.
(Ⅱ)不妨假設(shè)x
1≥x
2,而a<-1,由(Ⅰ)知在(0,+∞)單調(diào)減少,
從而?x
1,x
2∈(0,+∞),|f(x
1)-f(x
2)|≥4|x
1-x
2|
等價于?x
1,x
2∈(0,+∞),f(x
2)+4x
2≥f(x
1)+4x
1①
令g(x)=f(x)+4x,則

①等價于g(x)在(0,+∞)單調(diào)減少,即

.
從而

故a的取值范圍為(-∞,-2].(12分)
點評:本小題主要考查函數(shù)的導(dǎo)數(shù),單調(diào)性,極值,不等式等基礎(chǔ)知識,考查綜合利用數(shù)學知識分析問題、解決問題的能力.